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Vector field flow over upper surface of sphere

  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate the flow over the upper surface of sphere ##x^2+y^2+z^2=1## with normal vector pointed away from origin. Vector field is given as ##\vec{F}=(z^2x,\frac{1}{3}y^3+tan(z),x^2z+y^2)##


    2. Relevant equations
    Gaussian law: ##\int \int _{\partial \Sigma }\vec{F}d\vec{S}=\int \int \int _{\Sigma }\nabla \vec{F}dV##


    3. The attempt at a solution

    The idea here is to see that ##\int \int _{\partial \Sigma }\vec{F}d\vec{S}+\int\int_O \vec{F}d\vec{S}=\int \int \int _{\Sigma }\nabla \vec{F}dV##

    Where I marked the circle (plane) at ##z=0##.

    Well, I know how to solve this problem yet I have some problems, let me show you where... Let's start with the integral on the right side. We will need spherical coordinates ##x=rsin\theta cos\phi ##, ##y=rsin\theta sin\phi ## and ##z=rcos\phi ##. The integral is than:

    ##\int \int \int _{\Sigma }\nabla \vec{F}dV)=\int_{0}^{2\pi }d\phi \int_{0}^{\pi /2}d\theta\int_{0}^{1}(r^2sin^2\theta cos^2\phi +r^2sin^2\theta sin^2\phi +r^2cos\theta )r^2sin\theta dr##

    one ##r^2## comes from Jacobian determinant.

    Let's forget about all the ##\theta ## and ##\phi ##. Integration by r is obviously ##\frac{r^5}{5}##.

    OK.

    Now the left side of equation ##\int \int _{\partial \Sigma }\vec{F}d\vec{S}+\int\int_O \vec{F}d\vec{S}=\int \int \int _{\Sigma }\nabla \vec{F}dV##, specifically integral over surface ##O##.

    Using polar coordinates ##x=rcos\phi ##, ##y=rsin\phi ## and obviously ##z=0##. The parameterization is than ##r(r,\phi )=(rcos\phi ,rsin\phi , 0)## and ##r_r \times r_{\phi}=(0,0,r)##. Therefore the integral over ##O## for ##z=0##:

    ##\int\int_O \vec{F}d\vec{S}=-\int_{0}^{2\pi }d\phi \int_{0}^{1}(0,something,r^2sin^2 \phi)(0,0,r)dr##

    Jacobian is included in normal vector. Sorry, for using "something" instead of calculating it.

    Now let's again forget about all the angles and integrate by ##r##, the integral is than ##\frac{r^4}{4}##..

    My question here: WHAT IS WRONG? The units don't match!
     
    Last edited: Jan 9, 2014
  2. jcsd
  3. Jan 9, 2014 #2

    BruceW

    User Avatar
    Homework Helper

    true, there is an ##r^5## term on the right hand side and a ##r^4## term on the left hand side. But this is not a problem. In this situation, ##r## does not have the dimensions of length. It is dimensionless. Look at the original formula given: ##x^2+y^2+z^2=1## This tells you all you need to know about dimensions. One other thing to keep in mind, is that yes, there is an ##r^4## term on the left hand side, but once you finish the integration, you will use the limit (which is just a number). So you just get numbers on the left hand side and right hand side. No problem.
     
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